Question

一元二次方程,已知m,n是方程x²-2x-1=0的两个实数根,则3m²-n²-8m+1=?

Answer 1

用韦达定理,如果一元二次方程 aX^2+bX+C=0(a不等于0)

方程的两根X1,X2和方程的系数a,b,c就满足X1+X2=-(b/a）,X1*X2=c/a

m+n==2 m*n=-1

所以(m+n)^2=m^2+n^2+2mn=4

m^2+n^2=4-2mn=6

3m^2-n^2-8m+1

=3m^2+3n^2-4n^2-8m+1

=3(m^2+n^2)-4n^2-8m+1

=12-4n^2-8m+1

因为n是一元二次方程x2-2x-1=0的根

所以n^2-2n-1=0

n^2=2n+1

所以3m^2-n^2-8m+1

=12-4n^2-8m+1

=12-4(2n+1）－8m+1

=12－8n－4-8m+1

=12-8(m+n)-3=12-16-3=-7